Resposta:

[tex]\textsf{Leia abaixo}[/tex]

Explicação passo a passo:

[tex]\textsf{reta suporte de AB}[/tex]

[tex]\mathsf{m = \dfrac{\Delta_Y}{\Delta_X} = \dfrac{y_B - y_A}{x_B - x_A} = \dfrac{3 - (-3)}{-1 -(-4)} = \dfrac{3 + 3}{-1 + 4} = \dfrac{6}{3} = 2}[/tex]

[tex]\mathsf{y - y_0 = m(x - x_0)}[/tex]

[tex]\mathsf{y - 3 = 2(x - (-1))}[/tex]

[tex]\mathsf{y - 3 = 2x + 2}[/tex]

[tex]\mathsf{2x - y + 5 = 0}[/tex]

[tex]\textsf{reta suporte de H}[/tex]

[tex]\mathsf{m'.m'' = -1}[/tex]

[tex]\mathsf{m'.2 = -1}[/tex]

[tex]\mathsf{m' = -\dfrac{1}{2}}[/tex]

[tex]\mathsf{y - y_0 = m(x - x_0)}[/tex]

[tex]\mathsf{y - (-4) = -\dfrac{1}{2}(x - 3)}[/tex]

[tex]\mathsf{2y + 8= -x + 3}[/tex]

[tex]\mathsf{x + 2y + 5 = 0}[/tex]

[tex]\begin{cases}\mathsf{\mathsf{2x - y + 5 = 0}}\\\mathsf{x + 2y + 5 = 0}\end{cases}[/tex]

[tex]\mathsf{x + 2(2x + 5) = -5}[/tex]

[tex]\mathsf{x + 4x + 10 = -5}[/tex]

[tex]\mathsf{5x = -15}[/tex]

[tex]\mathsf{x = -3}[/tex]

[tex]\mathsf{-6 - y + 5 = 0}[/tex]

[tex]\mathsf{y = -1}[/tex]

[tex]\mathsf{H(-3;-1)}[/tex]

[tex]\mathsf{d_{CH} = \sqrt{(x_C - x_H)^2 + (y_C - y_H)^2}}[/tex]

[tex]\mathsf{d_{CH} = \sqrt{(3 - (-3))^2 + (-4 - (-1))^2}}[/tex]

[tex]\mathsf{d_{CH} = \sqrt{(3 + 3)^2 + (-4 + 1)^2}}[/tex]

[tex]\mathsf{d_{CH} = \sqrt{(6)^2 + (-3)^2}}[/tex]

[tex]\mathsf{d_{CH} = \sqrt{36 + 9}}[/tex]

[tex]\mathsf{d_{CH} = \sqrt{45}}[/tex]

[tex]\mathsf{d_{CH} = \sqrt{3^2.5}}[/tex]

[tex]\mathsf{d_{CH} = 3\sqrt{5}}[/tex]

[tex]\boxed{\boxed{\mathsf{d_{CH} = 6,7}}}[/tex]